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If $(1 -x + 2x^2)^n$ = $a_0 + a_1x + a_2x^2+..... a_{2n}x^{2n}$ , $n \in N$ , $x \in R$ and $a_0$ , $a_2$ and $a_1$ are in $A$ . $P$ .,then there exists
exactly two values of $n$
exactly one value of $n$
exactly three values of $n$
no value of $n$
Solution
$\left(1-\mathrm{x}+2 \mathrm{x}^{2}\right)^{\mathrm{n}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots$ …..$(1)$
$\mathrm{x}=0 \Rightarrow \mathrm{a}_{0}=1$
Diff. $(1)$
$n\left(1-x+2 x^{2}\right)^{n-1}(-1+4 x)$
$=a_{1}+2 a_{2} x+\ldots \ldots \ldots$ ………$(2)$
put $\mathrm{x}=0$
$-\mathrm{n}=\mathrm{a}_{1}$
Diff. $(2)$
${\mathrm{n}(\mathrm{n}-1)\left(1-\mathrm{x}+2 \mathrm{x}^{2}\right)^{\mathrm{n}-2}(-1+4 \mathrm{x})^{2}} $
${+\mathrm{n}\left(1-\mathrm{x}+2 \mathrm{x}^{2}\right)^{\mathrm{n}-1}(4)} $
${\quad=2 \mathrm{a}_{2}+\ldots \ldots \ldots}$ ………$(3)$
Put $\mathrm{x}=0$
$\mathrm{n}^{2}-\mathrm{n}+4 \mathrm{n}=2 \mathrm{a}_{2}$
$2 \mathrm{a}_{2}=\mathrm{n}^{2}+3 \mathrm{n}$
$\mathrm{a}_{1}+\mathrm{a}_{0}=2 \mathrm{a}_{2}$
$\Rightarrow-\mathrm{n}+1=\mathrm{n}^{2}+3 \mathrm{n}$
$\mathrm{n}^{2}+4 \mathrm{n}-1=0$
$\Rightarrow \mathrm{n} \notin \mathrm{N}$
